Problem: Agent Hunt transferred classified files from the CIA mainframe onto his flash drive. The drive had some files on it before the transfer, and the transfer happened at a rate of $4.4$ megabytes per second. After $32$ seconds, there were $384$ megabytes on the drive. The drive had a maximum capacity of $1000$ megabytes. How full was the drive when the transfer began?
Explanation: The files were transferred at a rate of $4.4$ megabytes per second, so $4.4T$ megabytes were transferred in $T$ seconds. The size of the files on the drive is comprised of the files that were on the drive before the transfer began and the files that were transferred. We can express this with the equation $S=A+4.4T$, where: $S$ represents the size of the files on the drive at a given time (in megabytes) $A$ represents the size of the files that were on the drive before the transfer began (in megabytes) $T$ represents the time (in seconds) We want to find $A$, so let's first solve the equation for $A$ : $ \begin{aligned}S&=A+4.4T\\ A&=S-4.4T\end{aligned}$ Now, we know that after $32$ seconds $(T={32})$, there were $384$ megabytes on the drive $(S={384})$. Let's plug these values into the equation to find the value of $A$. $ A={384}-4.4\cdot{32}=243.2$ Therefore, when the transfer began, the drive had $243.2$ megabytes on it. To find how long it took the drive to be completely full, we can plug $S=1000$ into the equation and solve for $T$. $ \begin{aligned}243.2&=1000-4.4T\\ 4.4T&=756.8\\ T&=172\end{aligned}$ When the transfer began, the drive had $243.2$ megabytes on it. It took the drive $172$ seconds to be completely full.